How do you find the area between #f(x)=sqrt(3x)+1, g(x)=x+1#?

Answer 1

The area is #3/2# square units.

Start by finding the intersection points.

#sqrt(3x) + 1 = x + 1#
#sqrt(3x) = x#
#(sqrt(3x))^2 = x^2#
#3x = x^2#
#x^2 - 3x = 0#
#x(x - 3) = 0#
#x = 0 and 3#
These will be our bounds of integration. Next, we use a test point in #[0, 3]# to determine which function is higher up.
Let the test point be #x = 2#.
#f(2) = sqrt(6) + 1 ~~ 3.449#
#g(2) = 2 + 1 = 3#
Therefore, #f(x)# will be above #g(x)#. We can now find an expression for the area:
#A = int_0^3 f(x) - g(x)dx#
#A = int_0^3 sqrt(3x) + 1 - (x + 1)dx#
#A = int_0^3 sqrt(3x) + 1 - x - 1 dx#
#A = int_0^3 sqrt(3x) - x dx#
#A = int_0^3 sqrt(3)sqrt(x) - xdx#
#A = [(2sqrt(3))/3x^(3/2) - 1/2x^2]_0^3#
#A = 2/3sqrt(3)(3)^(3/2) - 1/2(3)^2 - (2/3sqrt(3)(0^(3/2) - 1/2(0)^2))#
#A = 6 - 9/2#
#A = 3/2" "u^2#

Hopefully this helps!

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Answer 2

To find the area between the curves (f(x) = \sqrt{3x} + 1) and (g(x) = x + 1), you first need to determine the points of intersection between the two curves. Set (f(x) = g(x)) and solve for (x). Once you have the intersection points, integrate the absolute difference between the two functions over the interval between those points to find the area between the curves.

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Answer 3

To find the area between the curves ( f(x) = \sqrt{3x} + 1 ) and ( g(x) = x + 1 ), follow these steps:

  1. Determine the points of intersection between the two curves by setting them equal to each other and solving for ( x ).
  2. Determine which curve is above the other in the region of interest. This will help determine the limits of integration.
  3. Set up the integral to find the area between the curves by subtracting the lower curve from the upper curve and integrating with respect to ( x ) over the appropriate interval.

Let's go through these steps:

  1. Set ( \sqrt{3x} + 1 = x + 1 ) and solve for ( x ) to find the points of intersection: [ \sqrt{3x} = x ] [ 3x = x^2 ] [ x^2 - 3x = 0 ] [ x(x - 3) = 0 ]

So, ( x = 0 ) and ( x = 3 ).

  1. To determine which curve is above the other in the region of interest (from ( x = 0 ) to ( x = 3 )), compare the values of ( f(x) ) and ( g(x) ) in this interval.

[ f(x) = \sqrt{3x} + 1 ] [ g(x) = x + 1 ]

For ( x = 0 ): [ f(0) = \sqrt{3(0)} + 1 = 1 ] [ g(0) = 0 + 1 = 1 ]

For ( x = 3 ): [ f(3) = \sqrt{3(3)} + 1 = 3 ] [ g(3) = 3 + 1 = 4 ]

Since ( f(x) ) is above ( g(x) ) in the interval ( 0 \leq x \leq 3 ), we subtract ( g(x) ) from ( f(x) ) in the integral.

  1. Set up the integral to find the area between the curves: [ \text{Area} = \int_{0}^{3} (\sqrt{3x} + 1) - (x + 1) , dx ]

[ = \int_{0}^{3} (\sqrt{3x} - x) , dx ]

Now, you can integrate this expression over the interval ( 0 \leq x \leq 3 ) to find the area between the curves.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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